Problem: Find $\dfrac{d}{dx}\ln(3+x-x^3)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1-3x^2}{3+x-x^3}$ (Choice B) B $\dfrac{1}{3+x-x^3}$ (Choice C) C $\dfrac{1-3x^2}{\ln(x)(3+x-x^3)}$ (Choice D) D $\dfrac{1}{1-3x^2}$
Explanation: $\ln(3+x-x^3)$ is a logarithmic function, but its argument isn't simply $x$. Therefore, it's a composite function. In other words, suppose $u(x)=3+x-x^3$, then $\ln(3+x-x^3)=\ln\Bigl(u(x)\Bigr)$. $\dfrac{d}{dx}\ln(3+x-x^3)$ can be found using the following identity: $\dfrac{d}{dx}\left[\ln\Bigl(u(x)\Bigr)\right]=\dfrac{u'(x)}{u(x)}$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\ln(3+x-x^3) \\\\ &=\dfrac{d}{dx}\ln\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=3+x-x^3} \\\\ &=\dfrac{u'(x)}{u(x)} \\\\ &=\dfrac{1-3x^2}{3+x-x^3}&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\ln(3+x-x^3)=\dfrac{1-3x^2}{3+x-x^3}$.